Answer:
[tex]y = (x + \frac{1}{2} )^{2} + \frac{7}{4}[/tex]
Step-by-step explanation:
[tex]y = {x}^{2} + x + 2[/tex]
We can covert the standard form into the vertex form by either using the formula, completing the square or with calculus.
[tex]y = a(x - h)^{2} + k[/tex]
The following equation above is the vertex form of Quadratic Function.
Vertex — Formula
[tex]h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{4a} [/tex]
We substitute the value of these terms from the standard form.
[tex]y = a {x}^{2} + bx + c[/tex]
[tex]h = - \frac{1}{2(1)} \\ h = - \frac{ 1}{2} [/tex]
Our h is - 1/2
[tex]k = \frac{4(1)(2) - ( {1})^{2} }{4(1)} \\ k = \frac{8 - 1}{4} \\ k = \frac{7}{4} [/tex]
Our k is 7/4.
Vertex — Calculus
We can use differential or derivative to find the vertex as well.
[tex]f(x) = a {x}^{n} [/tex]
Therefore our derivative of f(x) —
[tex]f'(x) = n \times a {x}^{n - 1} [/tex]
From the standard form of the given equation.
[tex]y = {x}^{2} + x + 2[/tex]
Differentiate the following equation. We can use the dy/dx symbol instead of f'(x) or y'
[tex]f'(x) = (2 \times 1 {x}^{2 - 1} ) + (1 \times {x}^{1 - 1} ) + 0[/tex]
Any constants that are differentiated will automatically become 0.
[tex]f'(x) = 2 {x}+ 1[/tex]
Then we substitute f'(x) = 0
[tex]0 =2x + 1 \\ 2x + 1 = 0 \\ 2x = - 1 \\x = - \frac{1}{2} [/tex]
Because x = h. Therefore, h = - 1/2
Then substitute x = -1/2 in the function (not differentiated function)
[tex]y = {x}^{2} + x + 2[/tex]
[tex]y = ( - \frac{1}{2} )^{2} + ( - \frac{1}{2} ) + 2 \\ y = \frac{1}{4} - \frac{1}{2} + 2 \\ y = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} \\ y = \frac{7}{4} [/tex]
Because y = k. Our k is 7/4.
From the vertex form, our vertex is at (h,k)
Therefore, substitute h = -1/2 and k = 7/4 in the equation.
[tex]y = a {(x - h)}^{2} + k \\ y = (x - ( - \frac{1}{2} ))^{2} + \frac{7}{4} \\ y = (x + \frac{1}{2} )^{2} + \frac{7}{4} [/tex]
